Problem: The equation of a circle $C$ is $x^2+y^2+6x+4y-36 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Answer: To find the equation in standard form, complete the square. $(x^2+6x) + (y^2+4y) = 36$ $(x^2+6x+9) + (y^2+4y+4) = 36 + 9 + 4$ $(x+3)^{2} + (y+2)^{2} = 49 = 7^2$ Thus, $(h, k) = (-3, -2)$ and $r = 7$.